数据结构(英文版)课后习题答案

ProgrammingPrinciples 1THE GAME OF LIFEExercisesDetermine by hand calculation what will happen to each of the configurations shown in Figure overthe course of five generations. [Suggestion: Set up the Life configuration on a checkerboard. Use onecolor of checkers for living cells in the current generation and a second color to mark those that will beborn or die in the next generation.]Answer(a)Figure remains stable.(b)(c)(d)Figure is stable.12 Chapter 1 _ Programming Principles(e)(f)Figure repeats itself.(g)(h)(i)Figure repeats itself.(j)(k)(l)Figure repeats itself.Section _ Programming Style 3PROGRAMMING STYLEExercisesE1. What classes would you define in implementing the following projects What methods would your classespossess(a) A program to store telephone numbers.Answer The program could use classes called Phone_book and Person. The methods for a Phone_bookobject would include look_up_name, add_person, remove_person. The methods for a Personobject would include Look_up_number. Additional methods to initialize and print objects ofboth classes would also be useful.(b) A program to play Monopoly.Answer The program could use classes called Game_board, Property, Bank, Player, and Dice. In additionto initialization and printing methods for all classes, the following methods would be useful. Theclass Game_board needs methods next_card and operate_jail. The class Property needs methodschange_owner, look_up_owner, rent, build, mortgage, and unmortgage. The class Bank needsmethods pay and collect. The class Player needs methods roll_dice, move_location, buy_propertyand pay_rent. The class Dice needs a method roll.(c) A program to play tic-tac-toe.Answer The program could use classes called Game_board and Square. The classes need initializationand printing methods. The class Game_board would also need methods make_move andis_game_over. The class Square would need methods is_occupied, occupied_by, and occupy.(d) A program to model the build up of queues of cars waiting at a busy intersection with a traffic light.Answer The program could use classes Car, Traffic_light, and Queue. The classes would all need initializationand printing methods. The class Traffic_light would need additional methods change_statusand status. The class Queue would need additional methods add_car andremove_car.E2. Rewrite the following class definition, which is supposed to model a deck of playing cards, so that itconforms to our principles of style.class a { Y1 lists the cards. */ public:a( );void Shuffle( ); thing d( ); Given the declarationsint a[n][n], i, j;where n is a constant, determine what the following statement does, and rewrite the statement to accomplishthe same effect in a less tricky way.for (i = 0; i < n; i..)for (j = 0; j < n; j..)a[i][j] = ((i . 1)/(j . 1)) * ((j . 1)/(i . 1));Answer This statement initializes the array a with all 0’s except for 1’s down the main diagonal. A lesstricky way to accomplish this initialization is:for (i = 0; i < n; i..)for (j = 0; j < n; j..)if (i == j) a[i][j] = 1;else a[i][j] = 0;E4. Rewrite the following function so that it accomplishes the same result in a less tricky way.void does_something(int &first, int &second){first = second first;second = second first;first = second . first;}Answer The function interchanges the values of its parameters:void swap(int &first, int &second)/* Pre: The integers first and second have been initialized.Post: The values of first and second have been switched. */{int temp = first;first = second;second = temp;}E5. Determine what each of the following functions does. Rewrite each function with meaningful variablenames, with better format, and without unnecessary variables and statements. (a) int calculate(int apple, int orange){ int peach, lemon;peach = 0; lemon = 0; if (apple < orange)peach = orange; else if (orange <= apple)peach = apple; else { peach = 17;lemon = 19; }return(peach);}Answer The function calculate returns the larger of its two parameters.int larger(int a, int b)/* Pre: The integers a and b have been initialized.Post: The value of the larger of a and b is returned. */{if (a < b) return b;return a;}Section _ Programming Style 5(b) For this part assume the declaration typedef float vector[max];float figure (vector vector1){ int loop1, loop4; float loop2, loop3;loop1 = 0; loop2 = vector1[loop1]; loop3 = ;loop4 = loop1; for (loop4 = 0;loop4 < max; loop4..) { loop1 = loop1 . 1;loop2 = vector1[loop1 1];loop3 = loop2 . loop3; } loop1 = loop1 1;loop2 = loop1 . 1;return(loop2 = loop3/loop2); }Answer The function figure obtains the mean of an array of floating point numbers. float mean(vector v)/* Pre: The vector v contains max floating point values.Post: The mean of the values in v is returned. */{float total = ;for (int i = 0; i < max; i..)total += v[i];return total/((float) max);}(c) int question(int &a17, int &stuff){ int another, yetanother, stillonemore;another = yetanother; stillonemore = a17;yetanother = stuff; another = stillonemore;a17 = yetanother; stillonemore = yetanother;stuff = another; another = yetanother;yetanother = stuff; }Answer The function question interchanges the values of its parameters.void swap(int &first, int &second)/* Pre: The integers first and second have been initialized.Post: The values of first and second have been switched. */{int temp = first;first = second;second = temp;}(d) int mystery(int apple, int orange, int peach){ if (apple > orange) if (apple > peach) if(peach > orange) return(peach); else if (apple < orange)return(apple); else return(orange); else return(apple); elseif (peach > apple) if (peach > orange) return(orange); elsereturn(peach); else return(apple); }Answer The function mystery returns the middle value of its three parameters.6 Chapter 1 _ Programming Principlesint median(int a, int b, int c)/* Pre: None.Post: Returns the middle value of the three integers a, b, c. */{if (a > b)if (c > a) return a; The following statement is designed to check the relative sizes of three integers, which you may assumeto be different from each other:if (x < z) if (x < y) if (y < z) c = 1; else c = 2; elseif (y < z) c = 3; else c = 4; else if (x < y)if (x < z) c = 5; else c = 6; else if (y < z) c = 7; elseif (z < x) if (z < y) c = 8; else c = 9; else c = 10;(a) Rewrite this statement in a form that is easier to read.Answerif (x < z)if (x < y) Find those that can never occur, and eliminate the redundant checks. Answer The impossible cases are shown in the remarks for the preceding programsegment. After theirremoval we have:if (x < z)if (x < y) Answerif ((x < y) && (y < z)) c = 1;else if ((x < z) && (z < y)) c = 2;else if ((y < x) && (x < z)) c = 3;else if ((z < x) && (x < y)) c = 6;else if ((y < z) && (z < x)) c = 7;else c = 8;E7. The following C++ function calculates the cube root of a floating-point number (by the Newton approximation),using the fact that, if y is one approximation to the cube root of x, thenz . 2y . x=y23cube roots is a closer approximation.float function fcn(float stuff){ float april, tim, tiny, shadow, tom, tam, square; int flag;tim = stuff; tam = stuff; tiny = ;if (stuff != 0) do {shadow = tim . tim; square = tim * tim;tom = (shadow . stuff/square); april = tom/;if (april*april * april tam > tiny) if (april*april*april tam< tiny) flag = 1; else flag = 0; else flag = 0;if (flag == 0) tim = april; else tim = tam; } while (flag != 1);if (stuff == 0) return(stuff); else return(april); }(a) Rewrite this function with meaningful variable names, without the extra variables that contribute nothingto the understanding, with a better layout, and without the redundant and useless statements.Answer After some study it can be seen that both stuff and tam play the role of the quantity x in theformula, tim plays the role of y, and tom and april both play the role of z. The object tiny is asmall constant which serves as a tolerance to stop the loop. The variable shadow is nothing but2y and square is y2 . The complicated two-line if statement checks whether the absolute valuejz3 xj is less than the tolerance, and the boolean flag is used then only to terminatethe loop.Changing all these variables to their mathematical forms and eliminating the redundant onesgives:const double tolerance = ;double cube_root(double x) y . x/(y * y))/;y = z;} while (z * z * z x > tolerance || x z * z * z > tolerance);return z;}(b) Write a function for calculating the cube root of x directly from the mathematical formula, by startingwith the assignment y = x and then repeatingy = (2 * y . (x/(y * y)))/3until abs(y * y * y x) < .8 Chapter 1 _ Programming PrinciplesAnswer const double tolerance = ;double formula(double x) y . x/(y * y))/;} while (y * y * y x > tolerance || x y * y * y > tolerance);return y;}(c) Which of these tasks is easierAnswer It is often easier to write a program fromscratch than it is to decipher and rewrite a poorly writtenprogram.E8. The mean of a sequence of numbers is their sum divided by the count of numbers in the sequence. Thestatistics (population) variance of the sequence is the mean of the squares of all numbers in the sequence, minusthe square of the mean of the numbers in the sequence. The standard deviation is the square root of thevariance. Write a well-structured C++ function to calculate the standard deviation of a sequence of nfloating-point numbers, where n is a constant and the numbers are in an array indexed from 0 to n1,which is a parameter to the function. Use, then write, subsidiary functions to calculate the mean andvariance.Answer #include <>double variance(double v[], int n);double standard_deviation(double v[], int n) double mean(double v[], int n); double variance(double v[], int n)for (i = 0; i < n; i..)sum_squares += v[i] * v[i];temp = mean(v, n);return sum_squares/n temp * temp;}This function in turn requires another function to calculate the mean.double mean(double v[], int n)sum += v[i];return sum/n;}Section _ Programming Style 9E9. Design a program that will plot a given set of points on a graph. The input to the program will be a textfile, each line of which contains two numbers that are the x and y coordinates of a point to be plotted.The program will use a function to plot one such pair of coordinates. The details of the function involveplotting the specificmethod of plotting and cannot be written since they depend on the requirements of the plottingequipment, which we do not know. Before plotting the points the program needs to know the maximumand minimum values of x and y that appear in its input file. The program should therefore use anotherfunction bounds that will read the whole file and determine these four maxima and minima. Afterward,another function is used to draw and label the axes; then the file can be reset and the individual pointsplotted.(a) Write the main program, not including the functions.Answer #include <>#include ""#include ""#include ""int main(int argc, char *argv[]){ifstream file(argv[1]);if (file == 0) {cout << "Can not open input points file" << endl;cout << "Usage:\n\t plotter input_points " << endl;exit(1);}double xmax, xmin; Answer void bounds(ifstream &file, double &xmax, double&xmin,double &ymax, double &ymin)Answer void draw_axes(double xmax, double xmin,double ymax, double ymin)/* Pre: The parameters xmin, xmax, ymin, and xmax give bounds for the x and y co-ordinates.Post: Draws and labels axes according to the given bounds. */{}void plot(double x, double y)/* Pre: The parameters x and y give co-ordinates of a point.Post: The point is plotted to the ouput graph. */{}CODING, TESTING, AND FURTHER REFINEMENTExercisesE1. If you suspected that the Life program contained errors, where would be a good place to insert scaffoldinginto the main program What information should be printed outAnswer Since much of the program’s work is done in neighbor_count, a good place would be withinthe loops of the update method, just before the switch statement. The values of row, col, andneighbor_count could be printed.E2. Take your solution to Section , Exercise E9 (designing a program to plot a set of points), and indicategood places to insert scaffolding if needed.Answer Suitable places might be after draw_axes (printing its four parameters) and (with a very smalltest file to plot) after the call to plot (printing the coordinates of the point just plotted).E3. Find suitable black-box test data for each of the following:(a) A function that returns the largest of its three parameters, which are floating-point numbers.Answereasy values: .1; 2; 3., .2; 3; 1., .3; 2; 1..typical values: .0; 0:5;9:6., .1:3; 3:5; 0:4., .2:1;3:5;1:6..extreme values: .0; 0; 0., .0; 1; 1., .1; 0; 1., .0; 0; 1.(b) A function that returns the square root of a floating-point number.Answereasy values: 1, 4, 9, 16.typical values: , , .extreme value: .illegal values: , 9.Section _ Coding, Testing, and Further Refinement 11(c) A function that returns the least common multiple of its two parameters, which must be positive integers.(The least common multiple is the smallest integer that is a multiple of both parameters. Examples:The least common multiple of 4 and 6 is 12, of 3 and 9 is 9, and of 5 and 7 is 35.) Answereasy values: .3; 4., .4; 8., .7; 3..typical values: .7; 8., .189; 433., .1081; 1173..illegal values: .7;6., .0; 5., .0; 0., .1;1..(d) A function that sorts three integers, given as its parameters, into ascending order. Answereasy values: .5; 3; 2., .2; 3; 5., .3; 5; 2., .5; 2; 3., .1;2;3.extreme values: .1; 1; 1., .1; 2; 1., .1; 1; 2..typical values: .487;390; 20., .0; 589; 333..(e) A function that sorts an array a containing n integers indexed from 0 to n 1 into ascending order,where a and n are both parameters.Answer For the number n of entries to be sorted choose values such as 2, 3, 4 (easy values), 0, 1, maximumsize of a (extreme values), and 1 (illegal value). Test with all entries of a the same value, theentries already in ascending order, the entries in descending order, and the entries inrandomorder.E4. Find suitable glass-box test data for each of the following:(a) The statementif (a < b) if (c > d) x = 1; else if (c == d) x = 2;else x = 3; else if (a == b) x = 4; else if (c == d) x = 5;else x = 6;Answer Choose values for a and b, such as .1; 2., .2; 1., and .1; 1., so that each of a < b, a > b, and a ==b holds true. Choose three similar pairs of values forc and d, giving nine sets of test data.(b) The Life method neighbor_count(row, col).Answer Set row in turn to 1, maxrow, and any intermediate value, as well as 0 and maxrow . 1 (as illegalvalues). Choose col similarly. For each of the legal (row, col) pairs set up the Life object so thatthe number of living neighbors of (row, col) is each possible value between 0 and 8. Finally, makethe cell at (row, col) itself either living or dead. (This process gives 98 sets of test data, providedmaxrow and maxrow are each at least 3.)Programming ProjectsP1. Enter the Life program of this chapter on your computer and make sure that it works correctly.Answer The complete program is implemented in the life subdirectory for Chapter 1. #include "../../c/"#include ""#include "../../c/"#include ""int main() /*Pre: The user supplies an initial configuration of living cells.Post: The program prints a sequence of pictures showing the changes inthe configuration of living cells according to the rules forthe game of Life.Uses: The class Life and its methods initialize(), print(), andupdate(); the functions instructions(), user_says_yes().*/12 Chapter 1 _ Programming Principles{Life configuration;instructions();();();cout << "Continue viewing new generations " << endl;while (user_says_yes()) {();();cout << "Continue viewing new generations " << endl;}}const int maxrow = 20, maxcol = 60; int neighbor_count(int row, int col); };void Life::print()/*Pre: The Life object contains a configuration.Post: The configuration is written for the user.*/{int row, col;cout << "\nThe current Life configuration is:" <<endl;for (row = 1; row <= maxrow; row++) {for (col = 1; col <= maxcol; col++)if (grid[row][col] == 1) cout << '*';else cout << ' ';cout << endl;}cout << endl;}int Life::neighbor_count(int row, int col)/*Pre: The Life object contains a configuration, and the coordinatesrow and col define a cell inside its hedge.Post: The number of living neighbors of the specified cell is returned.*/{int i, j;int count = 0;for (i = row - 1; i <= row + 1; i++)for (j = col - 1; j <= col + 1; j++)count += grid[i][j]; count -= grid[row][col]; return count;}Section _ Coding, Testing, and Further Refinement 13void Life::update()/*Pre: The Life object contains a configuration.Post: The Life object contains the next generation of configuration. */{int row, col;int new_grid[maxrow + 2][maxcol + 2];for (row = 1; row <= maxrow; row++)for (col = 1; col <= maxcol; col++)switch (neighbor_count(row, col)) {case 2:new_grid[row][col] = grid[row][col]; break;case 3:new_grid[row][col] = 1; break;default:new_grid[row][col] = 0; }for (row = 1; row <= maxrow; row++)for (col = 1; col <= maxcol; col++)grid[row][col] = new_grid[row][col];}void Life::initialize()/*Pre: None.Post: The Life object contains a configuration specified by the user. */{int row, col;for (row = 0; row <= maxrow+1; row++)for (col = 0; col <= maxcol+1; col++)grid[row][col] = 0;cout << "List the coordinates for living cells." << endl;cout << "Terminate the list with the the special pair -1 -1" << endl;cin >> row >> col;while (row != -1 || col != -1) {if (row >= 1 && row <= maxrow)if (col >= 1 && col <= maxcol)grid[row][col] = 1;elsecout << "Column " << col << " is out of range." << endl;elsecout << "Row " << row << " is out of range." << endl;cin >> row >> col;}}void instructions()/*Pre: None.Post: Instructions for using the Life program have been printed.*/14 Chapter 1 _ Programming Principles{cout << "Welcome to Conway's game of Life." << endl;cout << "This game uses a grid of size "<< maxrow << " by " << maxcol << " in which" << endl;cout << "each cell can either be occupied by an organism or not." << endl;cout << "The occupied cells change from generation to generation" << endl;cout << "according to the number of neighboring cells which are alive."<< endl;}P2. Test the Life program with the examples shown in Figure .Answer See the solution to the exercise in Section .P3. Run the Life program with the initial configurations shown in Figure . Several of these go throughmany changes before reaching a configuration that remains the same or has predictable behavior.Answer This is a demonstration to be performed by computer.PROGRAM MAINTENANCEExercisesE1. Sometimes the user might wish to run the Life game on a grid smaller than 20_60.Determine how it ispossible to make maxrow and maxcol into variables that the user can set when the program is run. Tryto make as few changes in the program as possible.Answer The integers maxrow and maxcol should become data members of the class Life. The methodinitialize, must now ask for input of the two data members maxrow and maxcol. Upper boundsof 20 and 60 for these integers should be stored in new constants called maxrowbound andmaxcolbound. The amended file now takes the form.const int maxrowbound = 20, maxcolbound = 60;2][maxcolbound . 2];E2. One idea for speeding up the function Life :: neighbor_count(row, col) is to delete the hedge (the extrarows and columns that are always dead) from the arrays grid and new_grid. Then, when a cell is onthe boundary, neighbor_count will look at fewer than the eight neighboring cells, since some of theseare outside the bounds of the grid. To do this, the function will need to determine whether or not the cell(row, col) is on the boundary, but this can be done outside the nested loops, by determining, before theloops commence, the lower and upper bounds for the loops. If, for example, row is as small as allowed,then the lower bound for the row loop is row; otherwise, it is row 1. Determine, in terms of the size ofthe grid, approximately how many statements are executed by the original version of neighbor_countand by the new version. Are the changes proposed in this exercise worth making Section _ Program Maintenance 15Answer We need four if statements at the beginning of neighbor_count to determine whether (row, col)is on the boundary. This gives a total of 4 _ maxrow _ maxcol extra statements. If the cell is inthe first or last row or column, but not in a corner, then the nested loops would iterate 3 fewertimes. There are 8 such positions. With the cell in one of the 4 cornerpositions, the nested loops would iterate 5 fewer times. Hence the total number of statementssaved is4 _ maxrow _ maxcol . .2 _ maxrow . 2 _ maxcol 8..20:Thus this proposed change actually costs additional work, except for very small values of maxrowand maxcol.The modified function could be coded as follows.int Life :: neighbor_count(int row, int col)/* Pre: The Life object contains a configuration, and the coordinates row and col define a cellinside its hedge.Post: The number of living neighbors of the specified cell is returned. */{int i, j;int count = 0;int rowlow = row 1, rowhigh = row . 1,collow = col 1, colhigh = col . 1;if (row == 1) rowlow..;if (row == maxrow) rowhigh;if (col == 1) collow..;if (col == maxcol) colhigh;for (i = rowlow; i <= rowhigh; i..)for (j = collow; j <= colhigh; j..)count += grid[i][j]; Modify the Life function initialize so that it sets up the initial Life :: grid configuration by acceptingoccupied positions as a sequence of blanks and x’s in appropriate rows, rather than requiring the occupiedpositions to be entered as numerical coordinate pairs.Answer The following program includes all the changes for projects P1–P6. The changes required forProjects P7 and P8 are system dependent and have not been implemented.#include "../../c/"#include ""#include "../../c/"#include ""int main() /*Pre: The user supplies an initial configuration of living cells.Post: The program prints a sequence of pictures showing the changes in the configuration of living cells according to the rules forthe game of Life.Uses: The class Life and methods initialize(), print(), and update();the functions instructions(), user_says_yes().*/16 Chapter 1 _ Programming Principles{Life configuration;instructions();();(cout);bool command;do {();(cout);cout << "Continue viewing without changes " << endl;if (!(command = user_says_yes())) {cout << "Do you want to quit " << flush;if (user_says_yes()) command = false;else {cout << "Do you want help " << flush;if (user_says_yes()) instructions();cout << "Do you want to make any changes " << flush;if (user_says_yes()) ();command = true;}}} while (command);cout << "Do you want to save the final position to file " << flush;if (user_says_yes()) {char name[1000];cout << "Give the save file name: " << flush;cin >> name;ofstream f(name);(f);}}const int maxrow = 20, maxcol = 60; int neighbor_count(int row, int col); bool step_mode;};void Life::edit()/* Post: User has edited configuration, and/or step mode */{cout << "Do you want to switch the step mode " << flush;if (user_says_yes()) step_mode = !step_mode;cout << "Do you want to change the configuration " << flush;if (!user_says_yes()) return;Section _ Program Maintenance 17int row, col;cout << "List the coordinates for cells to change." << endl;cout << "Terminate the list with the the special pair -1 -1" << endl;cin >> row >> col;while (row != -1 || col != -1) {if (row >= 1 && row <= maxrow)if (col >= 1 && col <= maxcol)grid[row][col] = 1 - grid[row][col];elsecout << "Column " << col << " is out of range." << endl;elsecout << "Row " << row << " is out of range." << endl;cin >> row >> col;}}void Life::print(ostream &f)/*Pre: The Life object contains a configuration.Post: The configuration is written to stream f for the user.*/{int row, col;f << "The current Life configuration is:" << endl;for (row = 1; row <= maxrow; row++) {for (col = 1; col <= maxcol; col++)if (grid[row][col] == 1) f << '*';else f << ' ';f << endl;}f << endl;}int Life::neighbor_count(int row, int col)/*Pre: The Life object contains a configuration, and the coordinates row and col define a cell inside its hedge.Post: The number of living neighbors of the specified cell is returned. */{int i, j;int count = 0;for (i = row - 1; i <= row + 1; i++)for (j = col - 1; j <= col + 1; j++)count += grid[i][j]; count -= grid[row][col]; return count;}void Life::update()/*Pre: The Life object contains a configuration.Post: The Life object contains the next generation of configuration. */{int row, col;int new_grid[maxrow + 2][maxcol + 2];18 Chapter 1 _ Programming Principlesfor (row = 1; row <= maxrow; row++)for (col = 1; col <= maxcol; col++)switch (neighbor_count(row, col)) {case 2:new_grid[row][col] = grid[row][col]; break;case 3:if (step_mode &&grid[row][col] == 0)cout << "Vivify cell " << row << " " << col << endl;new_grid[row][col] = 1; break;default:if (step_mode && grid[row][col] == 1)cout << "Kill cell " << row << " " << col << endl;。